Question

A) How much heat (in kJ) is required to warm 13.0 g of ice, initially at -14.0 ∘C, to steam at 114.0 ∘C? The heat capacity of ice is 2.09 J/g⋅∘C and that of steam is 2.01 J/g⋅∘C

B) How much heat is evolved in converting 1.00 mol of steam at 130.0 ∘C to ice at -50.0 ∘C? The heat capacity of steam is 2.01 J/(g⋅∘C) and of ice is 2.09 J/(g⋅∘C).

Answer 1

a)

Q1 = m*Cp ice * (Tf – T1)

Q2 = m*LH ice

Q3 = m*Cp wáter * (Tb – Tf)

Q4 = m*LH vap

Q5 = m*Cp vap* (T2 – Tb)

Note that Tf = 0°C; Tb = 100°C, LH ice = 334 kJ/kg; LH water = 2264.76 kJ/kg.

Cp ice = 2.01 J/g°C ; Cp water = 4.184 J/g°C; Cp vapor = 2.030 kJ/kg°C

Then

Q1 = 13*2.01 * (0 – -14) = 365.82

Q2 = 13*334 = 4342

Q3 = 13*4.184 * (100 – 0) = 5439.2

Q4 = 13*2264.76 = 29441.88

Q5 = 13*2.03* (114– 100) = 369.46

QT = Q1+Q2+Q3+Q4+Q5 = 365.82+ 4342+5439.2+29441.88+369.46 = 39958.36 J

QT = 39.958 kJ

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